Work done on an object traveling down a slope

Click For Summary

Homework Help Overview

The discussion revolves around the concept of work done on an object traveling down a slope, particularly focusing on the relationship between gravitational potential energy and the work done by forces acting on the object. The original poster expresses confusion regarding Feynman's assertion that the work done by the force of constraint is zero, despite the presence of components in both the force and the direction of motion.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of work and the conditions under which the work done by the force of constraint is considered zero. Questions are raised about the relationship between the components of the force and the displacement vector.

Discussion Status

Some participants have provided insights into the reasoning behind the zero work done by the force of constraint, referencing the perpendicular nature of the force to the direction of motion. There is an ongoing exploration of the mathematical relationships involved, particularly concerning the dot product of vectors.

Contextual Notes

There is a noted emphasis on understanding the definitions and relationships between forces and motion in the context of gravitational potential energy, with participants questioning the assumptions made in the original statement.

Roo2
Messages
44
Reaction score
0

Homework Statement



I'm reading Feynman's lectures on physics to brush up on material I haven't looked at in a while, and I got confused by one of his statements. He claims that the change in potential energy of an object traveling down a slope can be calculated solely by the change in gravitational potential between the starting and ending point, irrespective of the force of constraint. While this seems intuitively correct to me, I'm confused by his explanation, in which he states that the work by gravity is nonzero (mgΔh) while the work done by the force of constraint is zero. To illustrate this, he provides the figure below, to which I have added an x-axis for reference:

work_zps315ca862.png


I'm confused because work is f*dr. Clearly, the object travels along the y-axis (I forgot to label, but assume perpendicular to x) and the force of gravity has an (entirely) y component, so f*dy is a nonzero number. However, the resultant force is shown to have an x component, and the direction of motion also has an x component, so f*dx is also nonzero. Why is the work done by the resultant force considered to be zero?

Homework Equations



W = f*dr
U(grav) = -G(m1m2/r^2) =~ mgh

The Attempt at a Solution



Described above.

Not a homework question per se, but I figured that it resembles a homework question more so than a "real" physics discussion in the general physics forum.

Thanks for any help!
 
Physics news on Phys.org
Feynman's statement is that the force of constraint does no work. The force of constraint is always perpendicular to the slope, whereas the direction of motion is always parallel to the slope. Hence, the force is always perpendicular to the displacement, and the work (their dot product) is zero.

Makes sense?
 
Yes and no. In the diagram above, the force of constraint has an x component, as does the direction of motion. Therefore, how can the dot product be zero?

edit: it can be zero if Fx * dx = -Fy * dy. But why does that equality hold?
 
Roo2 said:
Yes and no. In the diagram above, the force of constraint has an x component, as does the direction of motion. Therefore, how can the dot product be zero?

edit: it can be zero if Fx * dx = -Fy * dy. But why does that equality hold?

That relation between the components of the vectors F and dr MUST hold, precisely because these two vectors are always perpendicular to each other, and hence their dot product is always zero. The rest comes from the definition of the dot product:

$$\mathbf{F} \cdot d\mathbf{r} \equiv F_xdx + F_ydy = 0 $$

Edit: I use boldface to denote vectors, which is pretty standard.
 
A-ha! I get it. Thanks for showing me how to think about it!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K